Question

我想获取列表列表的名称。因此，我想使用列表元素列表（第2级）粘贴（或添加）列表元素（第一级）的名称。这应该用＆＃34;。＆＃34;来完成。对我来说重要的是使名称与列表列表中的顺序相同。这里有一个例子（注意我有更多的子列表，因为每个元素有9个）没有值：

GER_WF_groups <- list(A=list(fruits.2000=matrix(),nuts.2000=matrix(),veggies.2000=matrix(),
            fruits.2001=matrix(),nuts.2001=matrix(),veggies.2001=matrix(),
            fruits.2002=matrix(),nuts.2002=matrix(),veggies.2002=matrix()), 
     B=list(fruits.2000=matrix(),nuts.2000=matrix(),veggies.2000=matrix(),
            fruits.2001=matrix(),nuts.2001=matrix(),veggies.2001=matrix(),
            fruits.2002=matrix(),nuts.2002=matrix(),veggies.2002=matrix()), 
     C=list(fruits.2000=matrix(),nuts.2000=matrix(),veggies.2000=matrix(),
            fruits.2001=matrix(),nuts.2001=matrix(),veggies.2001=matrix(),
            fruits.2002=matrix(),nuts.2002=matrix(),veggies.2002=matrix()))

直到知道我使用复杂的方法来获得这个有序的名称，但它有效，见下文：

years_sapply2 <- sapply(GER_WF_groups,function(x) paste("BLUE", names(x) ,sep="."))
str(years_sapply2)
years_mat2 <- matrix(years_sapply2, nrow=9, byrow=F)
years_mat3 <- matrix(years_mat2, ncol=1, byrow=F)
years_vec <- years_mat3[,1]
BLUEyears <- years_vec[1:9]
GREENyears <- BLUEyears
GREYyears <- BLUEyears
GREENyearsN <- sub("BLUE", "GREEN", GREENyears)
GREYyearsN <- sub("BLUE", "GREY", GREYyears)
years_namesALL <- c(BLUEyears, GREENyearsN, GREYyearsN)

我的问题是：有更快/更简单的方法来完成这项工作吗？感谢

＆＃34; years_namesALL＆＃34;是预期的结果。

Answer 1

也许试试这个：

gsub("C","GREY",gsub("A","BLUE",gsub("B","GREEN",names(unlist(GER_WF_groups)))))

编辑（无耻地复制@Ananda Mahto对同一问题的编辑）：

gsub("C","GREY",gsub("A","BLUE",gsub("B","GREEN",names(unlist(GER_WF_groups,recursive=F)))))

但嵌套的gsub显然不优雅。

Answer 2

Vongo方法的变体如下：

names(unlist(setNames(GER_WF_groups, c("BLUE", "GREEN", "GREY"))))
##  [1] "BLUE.fruits.2000"   "BLUE.nuts.2000"     "BLUE.veggies.2000" 
##  [4] "BLUE.fruits.2001"   "BLUE.nuts.2001"     "BLUE.veggies.2001" 
##  [7] "BLUE.fruits.2002"   "BLUE.nuts.2002"     "BLUE.veggies.2002" 
## [10] "GREEN.fruits.2000"  "GREEN.nuts.2000"    "GREEN.veggies.2000"
## [13] "GREEN.fruits.2001"  "GREEN.nuts.2001"    "GREEN.veggies.2001"
## [16] "GREEN.fruits.2002"  "GREEN.nuts.2002"    "GREEN.veggies.2002"
## [19] "GREY.fruits.2000"   "GREY.nuts.2000"     "GREY.veggies.2000" 
## [22] "GREY.fruits.2001"   "GREY.nuts.2001"     "GREY.veggies.2001" 
## [25] "GREY.fruits.2002"   "GREY.nuts.2002"     "GREY.veggies.2002"

如果您只有一个级别的嵌套，您可以尝试：

names(unlist(setNames(GER_WF_groups, c("BLUE", "GREEN", "GREY")), recursive = FALSE))

R：如何按原始顺序获取列表名称？

2 个答案: